Find the area of the region enclosed by the inner loop of the curve.

$r=2+4\mathrm{sin}\theta$

Priscilla Johnston
2021-12-20
Answered

Find the area of the region enclosed by the inner loop of the curve.

$r=2+4\mathrm{sin}\theta$

You can still ask an expert for help

veiga34

Answered 2021-12-21
Author has **32** answers

Find the interval in which the inner loop exists.

For this we take$r=0$

$\Rightarrow 2+4\mathrm{sin}\theta =0$

$\Rightarrow \mathrm{sin}\theta =-\frac{12}{}$

$\Rightarrow \theta =\frac{7\pi}{6},\frac{11\pi}{6}$

So inner loop lies between the lines$\theta =\frac{7\pi}{6}$ and $\theta =\frac{11\pi}{6}$

Since curve is symmetric about the vertical polar axis. So we canf ind the area of the loop by making double the area between$\theta =\frac{7\pi}{6}$ and $\theta =\frac{3\pi}{2}$

Then Area of the loop is$A=2{\int}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}{\frac{12}{r}}^{2}d\theta$

Therefore

$A={\int}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}{(2+4\mathrm{sin}\theta )}^{2}d\theta$

$A=4{\int}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}{(1+2\mathrm{sin}\theta )}^{2}d\theta$

$=4{\int}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}(1+2(1-\mathrm{cos}2\theta )+4\mathrm{sin}\theta )d\theta$

$=4{\int}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}(3-2\mathrm{cos}2\theta +4\mathrm{sin}\theta )d\theta$

$=4[3\theta -\mathrm{sin}2\theta +4{(-\mathrm{cos}\theta ]}_{7\frac{\pi}{6}}^{3\frac{\pi}{2}}$

$=4[\frac{9\pi}{2}-\frac{7\pi}{2}-\frac{32}{\sqrt{3}}]$

$A=4\pi -6\sqrt{3}$

For this we take

So inner loop lies between the lines

Since curve is symmetric about the vertical polar axis. So we canf ind the area of the loop by making double the area between

Then Area of the loop is

Therefore

Archie Jones

Answered 2021-12-22
Author has **34** answers

The inner loop of the curve is bounded by the 2 points where r=0. Plug that in and you get sin? = -1/2, so ? = 210 and 330 degrees, or 7?/6 and 11?/6. So the integral needed is

$\int (2+4\mathrm{sin}\left(\theta \right))d\theta$

Apply the Sum Rule:

$=\int 2d\theta +\int 4\mathrm{sin}\left(\theta \right)dth\eta$
BSJ
$\int 2d\theta =2\theta$

$\int 4\mathrm{sin}\left(\theta \right)d\theta =-4\mathrm{cos}\left(\theta \right)$

$2\theta -4\mathrm{cos}\left(\theta \right)$

$=2\theta -4\mathrm{cos}\left(\theta \right)+C$

Compute the boundaries:${\int}_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(2+4\mathrm{sin}\left(\theta \right))d\theta =\frac{11\pi}{3}-2\cdot \sqrt{3}-(2\cdot \sqrt{3}+\frac{7\pi}{3})$

$=\frac{11\pi}{3}-2\cdot \sqrt{3}-(2\cdot \sqrt{3}+\frac{7\pi}{3})$

Simplify:

$=\frac{4\pi}{3}-4\sqrt{3}$

Apply the Sum Rule:

Compute the boundaries:

Simplify:

RizerMix

Answered 2021-12-29
Author has **438** answers

Very detailed answer. Thanks.

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